Трећи разред средње школе

Крамерово правило - примери 2

Решавање једначина задатих у облику детерминанте.

Задаци

Текст задатака објашњених у видео лекцији.

Пр.2)   Решити следећи систем једначина.

           $2x + y - z = 7$

           $5x - 4y + 7z = 1$

           $7x - 3y + 6z = 8$

Пр.2)

\[\begin{gathered}
{D_s} = \left| {\begin{array}{*{20}{c}}
2&1&{ - 1} \\
5&{ - 4}&7 \\
7&{ - 3}&6
\end{array}} \right|\begin{array}{*{20}{c}}
2 \\
5 \\
7
\end{array}\begin{array}{*{20}{c}}
{} \\
{} \\
{}
\end{array}\begin{array}{*{20}{c}}
1 \\
{ - 4} \\
{ - 3}
\end{array} = - 48 + 49 + 15 - 28 + 42 - 30 = 0 \hfill \\
\hfill \\
{D_x} = \left| {\begin{array}{*{20}{c}}
7&1&{ - 1} \\
1&{ - 4}&7 \\
8&{ - 3}&6
\end{array}} \right|\begin{array}{*{20}{c}}
7 \\
1 \\
8
\end{array}\begin{array}{*{20}{c}}
{} \\
{} \\
{}
\end{array}\begin{array}{*{20}{c}}
1 \\
{ - 4} \\
{ - 3}
\end{array} = - 168 + 56 + 3 - 32 + 147 - 6 = 0 \hfill \\
\hfill \\
{D_y} = \left| {\begin{array}{*{20}{c}}
2&7&{ - 1} \\
5&1&7 \\
7&8&6
\end{array}} \right|\begin{array}{*{20}{c}}
2 \\
5 \\
7
\end{array}\begin{array}{*{20}{c}}
{} \\
{} \\
{}
\end{array}\begin{array}{*{20}{c}}
7 \\
1 \\
8
\end{array} = 12 + 343 - 40 + 7 - 112 - 210 = 0\begin{array}{*{20}{c}}
{}&{}&{}
\end{array} \hfill \\
\hfill \\
{D_z} = \left| {\begin{array}{*{20}{c}}
2&1&7 \\
5&{ - 4}&1 \\
7&{ - 3}&8
\end{array}} \right|\begin{array}{*{20}{c}}
2 \\
5 \\
7
\end{array}\begin{array}{*{20}{c}}
{} \\
{} \\
{}
\end{array}\begin{array}{*{20}{c}}
1 \\
{ - 4} \\
{ - 3}
\end{array} = - 64 + 7 - 105 + 196 + 6 - 40 = 0 \hfill \\
\end{gathered} \]

${D_s} = {D_x} = {D_y} = {D_z} \Rightarrow $ неодређен систем. Датиј систем решавамо Гаусовим поступком.

\[\begin{gathered}
2x + y - z = 7 \hfill \\
5x - 4y + 7z = 1 \hfill \\
\underline {7x - 3y + 6z = 8} \hfill \\
2x + y - z = 7 \hfill \\
13x + 3z = 29 \hfill \\
\underline {13x + 3z = 29} \hfill \\
3z = 29 - 13x \hfill \\
z = \frac{{29 - 13x}}{3} \hfill \\
2x + y - \frac{{29 - 13x}}{3} = 7 \hfill \\
y = 7 - 2x + \frac{{29 - 13x}}{3} \hfill \\
y = \frac{{21 - 6x + 29 - 13x}}{3} \hfill \\
y = \frac{{ - 19x + 50}}{3} \hfill \\
\hfill \\
\left( {x,y,z} \right) = \left( {x,\frac{{50 - 19x}}{3},\frac{{29 - 13x}}{3}} \right) \hfill \\
\end{gathered} \]